Assume a set $S$ with $|S|$ entries. Indeed, $S$ is the set of lattice points inside a $k$-sphere. Assume $V=S\oplus S$ where $\oplus$ is the Minkowski sum of two sets. Do you know any lower bound on the cardinality of $V$?
Obviously, a very loose bound would be $|V|>|S|$ but I wonder if tighter bounds exist?
any hit is appreciated.
For simplicity let k-dimensional sphere is centered at the lattice point.
So we can do a linear transform of this lattice to $R^n$, where $n < k$ is dimension of the lattice, n-dimensional sphere which is a "slice" of k-dimensional sphere by n-dimensional plane containing the lattice will be transformed to n-dimensinal ellipsoid.
Thus, we have n-dimensinal ellipsoid E with center in $R^n$ and your set S is a set of integer points inside E: $$ E =\{x \in R^n: \frac{x^2_1}{a_1^2} + ... + \frac{x^2_n}{a_n^2} = 1\} $$
$$ S = \{(x_1,...,x_n) \in Z^n: \frac{x^2_1}{a_1^2} + ... + \frac{x^2_n}{a_n^2} \le 1\} $$
You want to find lower estimation of the number of elements in $V = S \oplus S$. Maybe try to estimate the number of elements in $V \setminus S$? We can consider "extreme radial" elements of S. It`s easy to see 2n such elements: $r_i = (0,...,0, \pm [a_i], 0,...,0)$, $i=1,...,n$ (for simplicity suppose that all $a_i \ge 1$).
As a duplications of "extreme radial" elements all $2r_i = r_i + r_i \in V \setminus S$. Hence, we slightly improve trivial lower estimation: $$|V| \ge |S| + 2n$$
If we calculate "extreme radial" elements of S more precisely we can improve the estimation.